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Theorem sspsstri 4067
Description: Two ways of stating trichotomy with respect to inclusion. (Contributed by NM, 12-Aug-2004.)
Assertion
Ref Expression
sspsstri ((𝐴𝐵𝐵𝐴) ↔ (𝐴𝐵𝐴 = 𝐵𝐵𝐴))

Proof of Theorem sspsstri
StepHypRef Expression
1 or32 925 . 2 (((𝐴𝐵𝐵𝐴) ∨ 𝐴 = 𝐵) ↔ ((𝐴𝐵𝐴 = 𝐵) ∨ 𝐵𝐴))
2 sspss 4064 . . . 4 (𝐴𝐵 ↔ (𝐴𝐵𝐴 = 𝐵))
3 sspss 4064 . . . . 5 (𝐵𝐴 ↔ (𝐵𝐴𝐵 = 𝐴))
4 eqcom 2744 . . . . . 6 (𝐵 = 𝐴𝐴 = 𝐵)
54orbi2i 912 . . . . 5 ((𝐵𝐴𝐵 = 𝐴) ↔ (𝐵𝐴𝐴 = 𝐵))
63, 5bitri 275 . . . 4 (𝐵𝐴 ↔ (𝐵𝐴𝐴 = 𝐵))
72, 6orbi12i 914 . . 3 ((𝐴𝐵𝐵𝐴) ↔ ((𝐴𝐵𝐴 = 𝐵) ∨ (𝐵𝐴𝐴 = 𝐵)))
8 orordir 929 . . 3 (((𝐴𝐵𝐵𝐴) ∨ 𝐴 = 𝐵) ↔ ((𝐴𝐵𝐴 = 𝐵) ∨ (𝐵𝐴𝐴 = 𝐵)))
97, 8bitr4i 278 . 2 ((𝐴𝐵𝐵𝐴) ↔ ((𝐴𝐵𝐵𝐴) ∨ 𝐴 = 𝐵))
10 df-3or 1089 . 2 ((𝐴𝐵𝐴 = 𝐵𝐵𝐴) ↔ ((𝐴𝐵𝐴 = 𝐵) ∨ 𝐵𝐴))
111, 9, 103bitr4i 303 1 ((𝐴𝐵𝐵𝐴) ↔ (𝐴𝐵𝐴 = 𝐵𝐵𝐴))
Colors of variables: wff setvar class
Syntax hints:  wb 205  wo 846  w3o 1087   = wceq 1542  wss 3915  wpss 3916
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2109  ax-9 2117  ax-ext 2708
This theorem depends on definitions:  df-bi 206  df-an 398  df-or 847  df-3or 1089  df-tru 1545  df-ex 1783  df-sb 2069  df-clab 2715  df-cleq 2729  df-clel 2815  df-ne 2945  df-v 3450  df-in 3922  df-ss 3932  df-pss 3934
This theorem is referenced by:  ordtri3or  6354  sorpss  7670  sorpssi  7671  funpsstri  34379
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